Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const double eps = 1e-9;
double s, a, b;
double l[MAXN], r[MAXN];
int h[MAXN];
int main() {
  scanf("%lf %lf %lf", &s, &a, &b);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf %lf", &l[i], &r[i]);
    h[i] = (r[i] - l[i]);
  }
  for (int i = 1; i <= n; i++) {
    h[i] += h[i - 1];
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    double x, y;
    scanf("%lf %lf", &x, &y);
    if (x == 3 && y == 1) y = 1;
    double hh = 0.0, zx = x + (a - x) * y / (y - s),
           ex = x + (b - x) * y / (y - s);
    int L = lower_bound(r + 1, r + n + 1, zx) - r;
    if (L <= n && l[L] < zx) {
      hh += r[L] - zx;
      L++;
    }
    int R = lower_bound(r + 1, r + n + 1, ex) - r;
    if (R <= n && l[R] <= ex && ex <= r[R] + eps) {
      hh += ex - l[R];
    }
    R--;
    if (R < 0) R = 0;
    if (L < 0) L = 0;
    hh += h[R] - h[L - 1];
    printf("%.15f\n", (y - s) * hh / y);
  }
}